Take the 2-minute tour ×
Physics Stack Exchange is a question and answer site for active researchers, academics and students of physics. It's 100% free, no registration required.

Possible Duplicate:
Why isn't the symmetric twin paradox a paradox?

Suppose there are two identical rockets, each carrying one of two identical clocks and one of two identical observers. The rockets lie on a straight line through space, face away from each other, and are completely motionless with respect to each other. The two rockets will use exactly the same flight plans except in different directions, like so:


               <**********************************************************---\
                                                                             |
     /----------------------------------A B----------------------------------/
     |
     \---**********************************************************>

The direction, power, and duration of the various uses of thrust will be identical. The actions of the observers are also completely identical. It seems that the experience of the observer in either rocket should be identical to the experience of the observer in the other rocket.

Also, the rocket do not accelerate at all while on the portions of their paths that are composed of asterisks.

Consider the portions of the paths composed of asterisks and the effect of time dilation. Because of the difference in velocities, the observer in either rocket will "perceive" the clock in the other rocket showing a time earlier than the clock in his own rocket. However, because the situation is completely symmetrical, the reality in either rocket should be the same; the clocks at any given moment read exactly the same thing. For arbitrarily long "asterisk paths", the difference between real time in either rocket and the time observed in that rocket by the the observer in the other rocket becomes arbitrarily large. (I'm not entirely certain about the conclusions in this paragraph.)

Now suppose that both rockets also carry pairs of missiles such that the four missiles are all identical. The passenger in rocket A hates the passenger in rocket B, so he does some calculations to figure out when to fire his missiles in order to kill the passenger in rocket B. (The passenger in rocket A is a nasty fellow.) He has decided to fire his two missiles in exactly opposite directions in order to prevent his rocket from being accelerated.

So, when he does this, what happens?

Will the observer in rocket B, who is being fired upon, perceive the two missiles leaving rocket A but no longer in rocket A? Will he perceive them being in rocket A but not leaving rocket A? Will he perceive them both still being in rocket A but also leaving it?

If he does not perceive them leaving rocket A, and if we assume that there is a sufficiently large "time lag" between rockets A and B, will the observer in rocket A observe a collision when the observer in rocket B does not observe a collision (and does not get hurt or killed)? Will the observer in rocket B have no physical possibility of seeing the missile missile coming but get killed by it anyway?

What happens?

share|improve this question
1  
All these "paradoxes" are trivially resolved by drawing a space-time picture. –  Ron Maimon Jun 12 '12 at 2:21
    
surely, we discuss simultaneity somewhere on this site. –  Jerry Schirmer Jun 19 '12 at 20:20
add comment

marked as duplicate by Martin Beckett, David Z Jul 20 '12 at 3:31

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

1 Answer

I would be glad to answer your question but I think I did not get your point so far.

But there is one thing I can say. This problem deal with objet that do accelerate (the missiles) therefore your question cannot be answered with special relativity alone.

share|improve this answer
1  
I don't know why people keep saying that you can't reason about acceleration using just special relativity. The entire question can be straightforwardly answered with a spacetime diagram and there's no difficulty representing rocket acceleration on such a diagram. –  Dan Piponi Jun 19 '12 at 21:45
    
Supporting documentation: math.ucr.edu/home/baez/physics/Relativity/SR/acceleration.html –  Dan Piponi Jun 19 '12 at 22:00
    
You are right Dan. Sorry. –  elamine Jun 20 '12 at 20:04
add comment

Not the answer you're looking for? Browse other questions tagged or ask your own question.